Area of Surface of Revolution
Example 1.
Rotate a curve that is a part of graph of
f ( x)  x 3
between x = 0 and x = 1, around x – axis and calculate the area of the surface made by
that rotation.
The idea of approximation is using circular strips (frustum), here is with two:
*****
Homework:
Prove that the area of circular frustum, as shown
is precisely
 Rr 
2 
 L
 2 
(1.1)
[See the appendix to this lecture below.]
*****
We shall use
Rr
 f  mi  where mi is the midpoint of the i th subdivision interval.
2
If we use a lot of these strips then the strips themselves are very similar to rectangles and
we get a better and better approximation as the number of strips increases
n
S    2 f  mi  
i 1
 xi  xi 1    yi  yi 1 
2
n
 2  f  mi  1   f   ti  
2
i 1
xi
2
n
   2 f  mi  
i 1
 xi    yi 
2
using Mean-Value Theorem (!)
where ti is a value in the i th subdivision interval.
and taking a limit
n
S  lim 2  f  mi 
n 
i 1
 dy 
1  i 
 dxi 
2
the upper
border of
variable of
integration
2
 dy 
xi  2
 x 1   dx  dx
the lower
3
border of
variable of
integration
1
1
 2  x3 1   3x 2  dx  2  x3 1  9 x 4 dx
2
0
0
and the rest is easy just by using substitution u  1  9 x
1
2  x3 1  9 x 4 dx 
0
In general, the idea is

10
u
18 
1/ 2
1
du 

4
10
u 3/ 2   3.563121852
1
27 
2
the upper
border of
variable of
integration
2
2
 dy 
 r ( x)  1   dx  dx
the lower
border of
variable of
integration
while the similar idea can be used in the case when x is expressed in the terms of y in
which case we simply replace the roles of x and y in the formula above. (Note: r is a
radius of a typical strip of revolution.)
Example 2.
Calculate the area of surface of revolution made by rotation of a graph of
f ( x)  x2  1 0  x  1 , about y-axis.
Now we use the general idea once again, radius is this time very simple
the upper
border of
variable of
integration
2
2
 dy 
2
 r ( x)  1   dx  dx  2 0 x  1   2 x  dx
the lower
1
border of
variable of
integration
1
 2  x  1  4 x dx 
2

0


6
5
u
4
1/2
du
using u  1  4 x 2
1
5
u 3/2 
1

6
5

5  1  5.3304135
Notice that we have used x as a variable of integration in our examples. The variable
here does not depend on axis of revolution, it is simply our choice of the integral
setup.
That is because the term
 x    y 
2
2
we used in our initial idea for the length
(L) of frustum is symmetric with respect to both variables.
Thus the alternative integral would be, as it will be addressed in the class,
the upper
border of
variable of
integration
2
0
 1 
 dx 
2
r
(
y
)

1

dy

2

y

1

1


 dy




 dy 
1
the lower
 2 y 1 
2
border of
variable of
integration
0
 2  y  1 
1
0
1
5
dy  2  y  dy
4
4
1
we get again the same result

6
5
and using u  y 
5
4

5  1  5.3304135
Homework: Check webassign.
Appendix:
Area of Frustum
Take a look at the following pictures (next page) of frustum, unfolded frustum, and
projection of frustum parallel to the base plane of the frustum (plane containing bigger
circle of the frustum).
The following is obvious,
x
L

r Rr
due to similar triangles. Thus we get
x
Lr
.
Rr
Now the area of the frustum is calculated as the difference of the areas of two circular
sectors,
A
 x  L   2 R  x  2r
2
2
  x  L   2 R  x  2r
 x  R  r    LR  Lr  LR  L  r  R  
and that is precisely (1.1) formula we needed to show.